Question: Factor completely. $128 -32 x + 2 x^2=$ 2
First, we take a common factor of $2$. $128 -32 x + 2 x^2=2(64 -16 x +x^2)$ Now, let's factor $64 -16 x +x^2$. Both $64$ and $x^2$ are perfect squares, since $64=({8})^2$ and $x^2=({x})^2$. Additionally, $16 x$ is twice the product of the roots of $64$ and $x^2$, since $16 x=2({8})({x})$. $64 -16 x +x^2 = ({8})^2 - 2({8})({x})+({x})^2$ So we can use the square of a difference pattern to factor: ${a}^2 - 2( a)( b)+ {b}^2 =({a} - {b})^2$ In this case, ${a}={8}$ and ${b}={x}$ : $ ({8})^2 - 2({8})({x})+({x})^2 =({8} -{x})^2$ $\begin{aligned} 128 -32 x + 2 x^2 &=2(64 -16 x +x^2) \\\\ &=2(8 - x)^2 \end{aligned}$ In conclusion, the complete factorization is $2(8 - x)^2$ Remember that you can always check your factorization by expanding it.